# Legendre's Condition

## Theorem

Let $y =\map y x$ be a real function, such that:

$\map y a=A,\quad \map y b=B$

Let $J\sqbrk y$ be a functional, such that:

$\displaystyle J\sqbrk y=\int_a^b \map F {x,y,y'}\rd x$

where

$F\in C^2\closedint a b$

with respect to all its variables, and $C$ stands for differentiability class.

Then a necessary condition for $J\sqbrk y$ to have a minimum for $y=\hat y$ is

$F_{y'y'}\big\vert_{y=\hat y}\ge 0\quad\forall x\in\closedint a b$.

## Proof

### Lemma 1

Let $y=\map y x$ be a real function, such that:

$\map y a=A,\quad\map y b=B$

Let $J\sqbrk y$ be a functional, such that:

$\displaystyle J\sqbrk y=\int_a^b \map F {x,y,y'}\rd x$

where

$F\in C^2\closedint a b$

with respect to all its variables.

Then

$\displaystyle\delta^2 J\sqbrk{y;h}=\int_a^b\paren{\map P {x,\map y x }h'^2+\map Q {x,\map y x}h^2}\rd x$

where

$\displaystyle P\paren{x,\map y x}=\frac 1 2 F_{y'y'},\quad \map Q {x,\map y x}=\frac 1 2 \paren {F_{yy}-\frac \d {\d x}F_{yy'} }$

### Lemma 2

Let $h$ be a real function such that:

$h\in C^1 \openint a b,\quad\map h a=0,\quad \map h b=0$

Let

$\displaystyle\delta^2 J\sqbrk{y;h}=\int_a^b \paren{\map P {x,\map y x}h'^2+\map Q {x,\map y x}h^2}\rd x$

where $P\in C^0\closedint a b$.

Then a necessary condition for

$\delta^2 J\sqbrk{y;h}\ge 0$

is

$\map P {x,\map y x}\ge 0\quad\forall x\in\closedint a b$

$\blacksquare$

## Source of Name

This entry was named for Adrien-Marie Legendre.